måndag 12 augusti 2013

Quantum Contradictions 7: Lithium Mystery Resolved

With the experience of the previous post we compute the ground state energy E_Li of Lithium as the energy of Li+ with two electrons filling a shell of radius 1/3 (= - 7.5 according to the previous post) plus the ionization energy E_Li3 of a third electron in an outer shell of radius d = 1 + 1/3 = 4/3, as follows:

E_Li3 = - 1/d + 1/ 2 x 1/1 = - 0.25

E_Li = - 7.5 - 0.25 = - 7.75

to be compared with the observed - 7.5.

We compare with the energy of a fictitious 1s3 ground state of Lithium, which is - 7.9 resulting from a radially symmetric (normalized) wave function with exponential decay exp( - 3r) in the distance r to the kernel for which the energy between two electrons is equal to 5/8 x 3 according to a standard text book computation.

The fictitious 1s3 state, which is viewed as unphysical because its energy is too small compared with the observed energy, in standard quantum mechanics is eliminated by Pauli's exclusion principle.

On the other hand, the state with two electrons filling an inner state of radius 1/3 with a third electron in an outer state of radius 4/3, has an energy in better agreement with observation and so may well be the true ground state. It can be motivated by the fact that the width of the outer third electron equals 1 from being attracted by an Li+ ion and thus is unable to enter into the inner shell of radius 1/3.

Similarly we obtain for Beryllium Be by adding two electrons in an outer shell of width 1/2 to Be++ with energy - 14 from two electrons in an inner shell of radius 1/4 (previous post) according to the following computation with d = 1/4 + 1/2 = 3/4:

E_Be34 = - 2 x 2/d + 2 x 1/2 x 1/(0.5)^2 + 1/2d = - 2/3

E_Be = - 14 - 2/3 = - 14.7

to be compared with observed - 14.8. The idea is again that the outer electrons are prevented from entering the inner shell because of too large width.

Next we will consider filling the outer shell with up to 8 electrons leading up to Neon and seek an explanation of the factor 8 as 2 x 4 with the 2 representing the 2 electrons of Helium, again without any Pauli exclusion principle (and thus no reference to spin).